We are working in $ X = \lbrace f : \mathbb{N} \rightarrow \mathbb{R} : \sum^{\infty}_{j = 1} |f(j)| < + \infty \rbrace $. We define $d_{l^1}(x_n, y_n) = \sum_{i \in \mathbb{N}} |x_i – y_i|$ and $d_{l^{\infty}}(x_n, y_n) = \sup_{i \in \mathbb{N}}|x_i – y_i|$.
My goal is to identify a sequence $ (f_{n})^{\infty}_{n=1} \in X$ which converges with respect to $d_{l^{\infty}}$ but doesn't converge with respect to $d_{l^{1}}$. I also want to show that a sequence in $X$ which converges with respect to $d_{l^1}$ must also converge with respect to $d_{l^{\infty}}$.
I'm a little confused about this because my understanding is that $X$ is just the $L^p$ space with $p = 1$ and I'm struggling to understand what it means for a sequence to converge in this space under a metric other than $d_{l^1}$.
Questions: what does it mean for a sequence to converge with respect to the $d_{l^{\infty}}$ metric in the $L^1$ space? When would a sequence in the $L^1$ space converge with respect to $d_{l^{\infty}}$ but fail to converge with respect to $d_{l^1}$?
EDIT: I think I actually found a sequence that works and proved it thanks to a friendly comment. See my answer and let me know if I made any mistakes in my proof.
Best Answer
The sequence in a previous answer is sufficient, but doesn't answer the other direction and I think the reasoning can be slightly more slick in this regard.
Note straight from the definitions that $\Vert \cdot \Vert_\infty \leq \Vert \cdot \Vert_1$, the former takes a $\sup$ over the same set of values where the latter takes a sum. This implies any convergent sequence in $\ell_1$ converges in $\ell_\infty$.
This additionally means that if $f_n$ converges to $f$ in $\ell_\infty$ and converges to $g$ in $\ell_1$ then $f=g$. That's because $\ell_\infty$ is a metric space, so has unique limits, and $\Vert f_n - g \Vert_\infty \leq \Vert f_n - g \Vert_1 \to 0$ i.e. $f_n$ also converges to $g$ in $\ell_\infty$.
Let $f_n = \chi_{\{m \leq n\}}\frac{1}{n}$ as in previously suggested answers. It converges to $0$ in $\ell_\infty$.
Assume that $f_n$ converges to $g$ in $\ell_1$. We must have $g=0$, but it is easy to check that $f_n$ does not converge to $0$ in $\ell_1$.